Help with linear transformation

Let $\displaystyle V$ be a vector space over $\displaystyle \mathbb{R}$, such that $\displaystyle dim(V)>1$.
Let $\displaystyle \varphi:V\rightarrow V$ be a linear transformation such that: $\displaystyle \varphi^2=-I$.
Prove that for every $\displaystyle v\neq 0$: $\displaystyle v,\varphi(v)$ are linear independent.

This is what I tried so far:
Let $\displaystyle v\in V$.
suppose $\displaystyle \alpha_1v+\alpha_2\varphi(v)=0$, I need to prove $\displaystyle \alpha_1=\alpha_2=0$.
since $\displaystyle \varphi$ is linear, and $\displaystyle \alpha_1v+\alpha_2\varphi(v)=0$, then also $\displaystyle \varphi(\alpha_1v+\alpha_2\varphi(v))=0$, and then:
$\displaystyle \varphi(\alpha_1v+\alpha_2\varphi(v))=0$
$\displaystyle \varphi(\alpha_1v)+\varphi(\alpha_2\varphi(v)))=0$
$\displaystyle \alpha_1\varphi(v)+\alpha_2\varphi^2(v)=0$ now since $\displaystyle \varphi^2=-I$:
$\displaystyle \alpha_1\varphi(v)-\alpha_2v=0$

Re: Help with linear transformation

Let $\displaystyle V$ be a vector space over $\displaystyle \mathbb{R}$, such that $\displaystyle dim(V)>1$.
Let $\displaystyle \varphi:V\rightarrow V$ be a linear transformation such that: $\displaystyle \varphi^2=-I$.
Prove that for every $\displaystyle v\neq 0$: $\displaystyle v,\varphi(v)$ are linear independent.

This is what I tried so far:
Let $\displaystyle v\in V$.
suppose $\displaystyle \alpha_1v+\alpha_2\varphi(v)=0$, I need to prove $\displaystyle \alpha_1=\alpha_2=0$.
since $\displaystyle \varphi$ is linear, and $\displaystyle \alpha_1v+\alpha_2\varphi(v)=0$, then also $\displaystyle \varphi(\alpha_1v+\alpha_2\varphi(v))=0$, and then:
$\displaystyle \varphi(\alpha_1v+\alpha_2\varphi(v))=0$
$\displaystyle \varphi(\alpha_1v)+\varphi(\alpha_2\varphi(v)))=0$
$\displaystyle \alpha_1\varphi(v)+\alpha_2\varphi^2(v)=0$ now since $\displaystyle \varphi^2=-I$:
$\displaystyle \alpha_1\varphi(v)-\alpha_2v=0$

but now I'm not sure what conclusion should I draw from this...

I can really use some guidance here.
Thanks in advanced!

Hi Stormey!

Suppose $\displaystyle v,\varphi(v)$ are linearly dependent.
Then there is some $\displaystyle t \in \mathbb R$ such that $\displaystyle \varphi(v) = t v$.